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A man invested ₹X in compound interest at 20% p.a. for two years, and he invested the same amount in simple interest for three years at 21% p.a. If the difference between the interests is ₹1900, then find X.
- 10000
- 12000
- 14000
- 8000
Correct answer: 10000
Solution
Compound interest for 2 years at 20% is \(X[(1.2)^2-1]=0.44X\). Simple interest for 3 years at 21% is \(X\times 21\times 3/100=0.63X\). Their difference is \(0.63X-0.44X=0.19X=1900\), so \(X=10000\).
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